-0.000 000 000 742 147 676 646 574 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 574(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 646 574(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 646 574| = 0.000 000 000 742 147 676 646 574


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 646 574.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 646 574 × 2 = 0 + 0.000 000 001 484 295 353 293 148;
  • 2) 0.000 000 001 484 295 353 293 148 × 2 = 0 + 0.000 000 002 968 590 706 586 296;
  • 3) 0.000 000 002 968 590 706 586 296 × 2 = 0 + 0.000 000 005 937 181 413 172 592;
  • 4) 0.000 000 005 937 181 413 172 592 × 2 = 0 + 0.000 000 011 874 362 826 345 184;
  • 5) 0.000 000 011 874 362 826 345 184 × 2 = 0 + 0.000 000 023 748 725 652 690 368;
  • 6) 0.000 000 023 748 725 652 690 368 × 2 = 0 + 0.000 000 047 497 451 305 380 736;
  • 7) 0.000 000 047 497 451 305 380 736 × 2 = 0 + 0.000 000 094 994 902 610 761 472;
  • 8) 0.000 000 094 994 902 610 761 472 × 2 = 0 + 0.000 000 189 989 805 221 522 944;
  • 9) 0.000 000 189 989 805 221 522 944 × 2 = 0 + 0.000 000 379 979 610 443 045 888;
  • 10) 0.000 000 379 979 610 443 045 888 × 2 = 0 + 0.000 000 759 959 220 886 091 776;
  • 11) 0.000 000 759 959 220 886 091 776 × 2 = 0 + 0.000 001 519 918 441 772 183 552;
  • 12) 0.000 001 519 918 441 772 183 552 × 2 = 0 + 0.000 003 039 836 883 544 367 104;
  • 13) 0.000 003 039 836 883 544 367 104 × 2 = 0 + 0.000 006 079 673 767 088 734 208;
  • 14) 0.000 006 079 673 767 088 734 208 × 2 = 0 + 0.000 012 159 347 534 177 468 416;
  • 15) 0.000 012 159 347 534 177 468 416 × 2 = 0 + 0.000 024 318 695 068 354 936 832;
  • 16) 0.000 024 318 695 068 354 936 832 × 2 = 0 + 0.000 048 637 390 136 709 873 664;
  • 17) 0.000 048 637 390 136 709 873 664 × 2 = 0 + 0.000 097 274 780 273 419 747 328;
  • 18) 0.000 097 274 780 273 419 747 328 × 2 = 0 + 0.000 194 549 560 546 839 494 656;
  • 19) 0.000 194 549 560 546 839 494 656 × 2 = 0 + 0.000 389 099 121 093 678 989 312;
  • 20) 0.000 389 099 121 093 678 989 312 × 2 = 0 + 0.000 778 198 242 187 357 978 624;
  • 21) 0.000 778 198 242 187 357 978 624 × 2 = 0 + 0.001 556 396 484 374 715 957 248;
  • 22) 0.001 556 396 484 374 715 957 248 × 2 = 0 + 0.003 112 792 968 749 431 914 496;
  • 23) 0.003 112 792 968 749 431 914 496 × 2 = 0 + 0.006 225 585 937 498 863 828 992;
  • 24) 0.006 225 585 937 498 863 828 992 × 2 = 0 + 0.012 451 171 874 997 727 657 984;
  • 25) 0.012 451 171 874 997 727 657 984 × 2 = 0 + 0.024 902 343 749 995 455 315 968;
  • 26) 0.024 902 343 749 995 455 315 968 × 2 = 0 + 0.049 804 687 499 990 910 631 936;
  • 27) 0.049 804 687 499 990 910 631 936 × 2 = 0 + 0.099 609 374 999 981 821 263 872;
  • 28) 0.099 609 374 999 981 821 263 872 × 2 = 0 + 0.199 218 749 999 963 642 527 744;
  • 29) 0.199 218 749 999 963 642 527 744 × 2 = 0 + 0.398 437 499 999 927 285 055 488;
  • 30) 0.398 437 499 999 927 285 055 488 × 2 = 0 + 0.796 874 999 999 854 570 110 976;
  • 31) 0.796 874 999 999 854 570 110 976 × 2 = 1 + 0.593 749 999 999 709 140 221 952;
  • 32) 0.593 749 999 999 709 140 221 952 × 2 = 1 + 0.187 499 999 999 418 280 443 904;
  • 33) 0.187 499 999 999 418 280 443 904 × 2 = 0 + 0.374 999 999 998 836 560 887 808;
  • 34) 0.374 999 999 998 836 560 887 808 × 2 = 0 + 0.749 999 999 997 673 121 775 616;
  • 35) 0.749 999 999 997 673 121 775 616 × 2 = 1 + 0.499 999 999 995 346 243 551 232;
  • 36) 0.499 999 999 995 346 243 551 232 × 2 = 0 + 0.999 999 999 990 692 487 102 464;
  • 37) 0.999 999 999 990 692 487 102 464 × 2 = 1 + 0.999 999 999 981 384 974 204 928;
  • 38) 0.999 999 999 981 384 974 204 928 × 2 = 1 + 0.999 999 999 962 769 948 409 856;
  • 39) 0.999 999 999 962 769 948 409 856 × 2 = 1 + 0.999 999 999 925 539 896 819 712;
  • 40) 0.999 999 999 925 539 896 819 712 × 2 = 1 + 0.999 999 999 851 079 793 639 424;
  • 41) 0.999 999 999 851 079 793 639 424 × 2 = 1 + 0.999 999 999 702 159 587 278 848;
  • 42) 0.999 999 999 702 159 587 278 848 × 2 = 1 + 0.999 999 999 404 319 174 557 696;
  • 43) 0.999 999 999 404 319 174 557 696 × 2 = 1 + 0.999 999 998 808 638 349 115 392;
  • 44) 0.999 999 998 808 638 349 115 392 × 2 = 1 + 0.999 999 997 617 276 698 230 784;
  • 45) 0.999 999 997 617 276 698 230 784 × 2 = 1 + 0.999 999 995 234 553 396 461 568;
  • 46) 0.999 999 995 234 553 396 461 568 × 2 = 1 + 0.999 999 990 469 106 792 923 136;
  • 47) 0.999 999 990 469 106 792 923 136 × 2 = 1 + 0.999 999 980 938 213 585 846 272;
  • 48) 0.999 999 980 938 213 585 846 272 × 2 = 1 + 0.999 999 961 876 427 171 692 544;
  • 49) 0.999 999 961 876 427 171 692 544 × 2 = 1 + 0.999 999 923 752 854 343 385 088;
  • 50) 0.999 999 923 752 854 343 385 088 × 2 = 1 + 0.999 999 847 505 708 686 770 176;
  • 51) 0.999 999 847 505 708 686 770 176 × 2 = 1 + 0.999 999 695 011 417 373 540 352;
  • 52) 0.999 999 695 011 417 373 540 352 × 2 = 1 + 0.999 999 390 022 834 747 080 704;
  • 53) 0.999 999 390 022 834 747 080 704 × 2 = 1 + 0.999 998 780 045 669 494 161 408;
  • 54) 0.999 998 780 045 669 494 161 408 × 2 = 1 + 0.999 997 560 091 338 988 322 816;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 646 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 574 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111